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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 16201. |
The length of the second diagonal of rhombus whose side is 5cm and one of the diagonal is 8cm is |
| Answer» 6 cm | |
| 16202. |
The 11th term of the AP -5,-5,2,0,5,2,....is(a)-20 (b)20(c)-30(d)30 |
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Answer» Question is wrong because this is not an A.P 2(4)+4 I think this is not any A.P. because their difference is not same you can find this by formula Tn= a+ ( n-1) d |
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| 16203. |
The list of numbers -10,-6,-2,2....is (A)an AP with d=4 (B)not an Ap |
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Answer» Hii Option A is right same Hi (A) yes it is an ap bcoz its common difference is in all terms |
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| 16204. |
Koi fb use karta h Kya |
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Answer» disgusting manner Noo Nooo No krti thi pehle ab nhi I\'d ka username send Karo |
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| 16205. |
What is the section formula? |
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Answer» X=m1x2+m2x1/m1+m2 and y=m2y1+m1y2/m1+m2 mx2+nx1/m+n , my2+ny1/m+n |
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| 16206. |
Can anyone knows about the section formula ch 7 cordinate geometry please define it if anyone knows |
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| 16207. |
7√2x-10x-4√2 |
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| 16208. |
If the sum of m term of AP is the same as the sum of its n term then show that (m+n)th term is zero |
| Answer» mam = nanm[a + (m - 1)d] = n [a + (n - 1)d]{tex} \\Rightarrow {/tex}\xa0ma + m2d - md = na + n2d - nd{tex} \\Rightarrow {/tex}\xa0a(m - n) + (m2 - n2)d - md + nd = 0{tex} \\Rightarrow {/tex}\xa0a(m - n) + (m - n) (m + n)d - (m - n)d = 0{tex} \\Rightarrow {/tex}\xa0(m - n) [a + (m + n - 1)d] = 0{tex} \\Rightarrow {/tex}\xa0a + (m + n - 1)d = 0{tex} \\Rightarrow {/tex}\xa0am+n = 0Hence proved. | |
| 16209. |
Find the value of k for which the lines ( k + 1 )x + 3ky +15 = 0 and 5x + ky + 5 = 0 are coincident |
| Answer» What is the answer | |
| 16210. |
Find a quadratic polynomial,whose zeroes are -3 and 2, respectively |
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| 16211. |
Cos 60 is equal to |
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Answer» Cos60°=1/2 1/2 Hey miss emoji..whats ur name 1/2 1/2 |
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| 16212. |
RD sharma class 10, Page no: 12.34, Question no: 67. |
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| 16213. |
If tan theta + 1upon tan theta=2 then find value of tan square theta + 1upon tan square theta |
| Answer» If tanA + 1/tanA = 2, find the value of tan²A + 1/tan²A.Squaring both sides, =) (tanA + 1/tanA)² = 2²=) tan²A + 1/tan²A + 2(tanA)(1/tanA) = 4=) tan²A + 1/tan²A + 2 = 4=) tan²A + 1/tan²A = 4-2=) tan²A + 1/tan²A = 2. | |
| 16214. |
2x+3y=5,4x+ky=10 find the value of k by infinity many solution |
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Answer» Equation 1: 2x + 3y = 5Equation 2: 4x + ky = 10Both the equations are in the form of :a1x + b1y = c1 & a2x + b2y = c2 where a1 & a2 are the coefficients of xb1 & b2 are the coefficients of yc1 & c2 are the constantsFor the system of linear equations to have infinitely many solutions we must havea1/a2 = b1/b2 = c1/c2 ………(i)According to the problem:a1 = 2, a2 = 4, b1 = 3, b2 = k, c1 = 5, c2 = 10Putting the above values in equation (i) and solving the extreme left and extreme right portion of the equality we get the value of a2/4 = 3/k⇒ 2k = 12 ⇒ k = 6The value of k for which the system of equations has infinitely many solution is k = 6 Given - infinitely many solution =2/4=3/k K=6 |
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| 16215. |
Why we shall use a value of pi as 22/7 |
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Answer» as we know Aryabhatta ,a great mathematician discovered the perimeter of circle in a rope phenomena by which a discovery of 22/7 asπ came It is easy for calculating except for some problems wherein you have to use 3.14 as the value of π For easy calculation. The original value of pi is 3.14 Because it\'s universal value of π |
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| 16216. |
how to get 500 / 500 in 10 th board exam |
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Answer» study with great lagan Do hardwork Focus and hardwork for any particular aim bring success to study hard in the exam Do hard work |
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| 16217. |
Prove /5 is irrational |
| Answer» let root 5 be rationalthen it must in the form of p/q [q is not equal to 0][p and q are co-prime]root 5=p/q=> root 5 × q = psquaring on both sides=> 5×q×q = p×p ------> 1p×p is divisible by 5p is divisible by 5p = 5c [c is a positive integer] [squaring on both sides ]p×p = 25c×c --------- > 2sub p×p in 15×q×q = 25×c×cq×q = 5×c×c=> q is divisble by 5thus q and p have a common factor 5there is a contradictionas our assumsion p &q are co prime but it has a common factorso\xa0√5 is an irrational | |
| 16218. |
For commerc we have to choose standard |
| Answer» Yes | |
| 16219. |
Is there any simpler way or trick to solve 8.4 exercise |
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Answer» No never but everything is possible try to solve tough questions like fourth ,fifth , sixth as much u can make it shorter by using each and every thing u know to solve aproblem U searh on youtube chanel in truemath |
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| 16220. |
Happt independence day ?????? |
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Answer» Haanji pata hai ap thik nhi ho aditi ji..take care nd get well soon☺ Gungun im not fine...& take care Same 2 youuu Same to you sis Happy Independence Day & Rakhsha Bandhan |
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| 16221. |
Prove that sin theeta - 2sin^3 theeta /2cos^3 theeta -cos theeta is equal to tan theeta |
| Answer» LHS = (sin{tex}\\theta{/tex}\xa0- 2sin3{tex}\\theta{/tex})= sin{tex}\\theta{/tex}(1 - 2sin2{tex}\\theta{/tex})RHS = (2cos3{tex}\\theta{/tex}\xa0- cos{tex}\\theta{/tex})tan{tex}\\theta{/tex}= cos{tex}\\theta{/tex}(2cos2{tex}\\theta{/tex}\xa0- 1){tex}\\frac { \\sin \\theta } { \\cos \\theta }{/tex}= [2(1 - sin2{tex}\\theta{/tex}) - 1)sin{tex}\\theta{/tex}= (2 - 2sin2{tex}\\theta{/tex}\xa0-1)sin{tex}\\theta{/tex}= (1 - 2sin2{tex}\\theta{/tex})sin{tex}\\theta{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = RHS{tex}\\therefore{/tex}\xa0(sin{tex}\\theta{/tex}\xa0- 2sin3{tex}\\theta{/tex}) = (2cos3{tex}\\theta{/tex}\xa0- cos{tex}\\theta{/tex})tan{tex}\\theta{/tex} | |
| 16222. |
If sec∅+tan∅=x find the value of sec∅ |
| Answer» Given:\xa0sec x + tan x = pAs we know,{tex}sec^2 x - tan^2 x = 1{/tex}{tex}\\therefore{/tex}\xa0{tex}(sec\\ x + tan\\ x)(sec\\ x - tan\\ x) = 1\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}p(sec\\ x - tan\\ x) = 1\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}sec\\ x - tan\\ x ={/tex}\xa0{tex}\\frac{1}{p}{/tex}Thus, we have{tex}sec\\ x + tan\\ x = p{/tex} and\xa0sec x - tan x =\xa0{tex}\\frac{1}{p}{/tex}{tex}\\Rightarrow{/tex}\xa0(sec x + tan x) + (sec x - tan x) = p +\xa0{tex}\\frac{1}{p}{/tex}\xa0and\xa0(sec x + tan x) - (sec x - tan x) = p -\xa0{tex}\\frac{1}{p}{/tex}{tex}\\Rightarrow{/tex}\xa02 sec x = p +\xa0{tex}\\frac{1}{p}{/tex}\xa0and 2 tan x = -{tex}\\frac{1}{p}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0sec x =\xa0{tex}\\frac{p^{2}+1}{2 p}{/tex}\xa0and\xa0tan x =\xa0{tex}\\frac{p^{2}-1}{2 p}{/tex} | |
| 16223. |
If the HCF of 657 and963 is 657x +963x -15, find x |
| Answer» Given numbers are 657 and 963 .Here, 657 < 963\xa0By using Euclid\'s Division algorithmm , we get963 = (657 × 1) + 306Here , remainder = 306 .So, On taking 657 as new dividend and 306 as the new divisor and then apply Euclid\'s Division lemma, we get657 = (306 × 2) + 45Here, remainder = 45\xa0So, On taking 306\xa0as new dividend and 45\xa0as the new divisor and then apply Euclid\'s Division lemma, we get306 = (45 × 6) + 36Here, remainder = 36So, On taking 45\xa0as new dividend and 36\xa0as the new divisor and then apply Euclid\'s Division lemma, we get45 = (36 × 1) + 9Here, remainder = 9So, On taking 36\xa0as new dividend and 9\xa0as the new divisor and then apply Euclid\'s Division lemma, we get36 = (9 × 4) + 0Here , remainder = 0 and last divisor is 9.\xa0Hence, HCF of 657 and 963 = 9.∴ 9 = 657x + 963(-15)⇒ 9 = 657x - 14445⇒ 657x = 9 + 14445⇒ 657x = 14454⇒x = 14454/657⇒ x =22 | |
| 16224. |
3x-5y=49x=2x+5y |
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| 16225. |
(6x-4y)+(5x6y)Distance formula |
| Answer» D=t/s | |
| 16226. |
Good evening sir/mam may i know where from i got mathematics both code (41 & 241) syllabus |
| Answer» You can check the syllabus here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 16227. |
Commerce Vs math which one is better |
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Answer» Maths our future Maths Pcb is the best....with computer If you are weak in maths then it is better to take without math..no problem....but best combination is commerce with math Commerce without math ??? U mean commerce with maths vs commerce without maths? |
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| 16228. |
Hello Everyone.....say something....? |
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Answer» Hi... Bye Then M for he said ....mam simple my mom\'s makeup???????? Then i asked what is P for ......1 of the students said Teacher its Pubg.....modern students are great!!!!????????? Btao..kya btana h Aaaj bccho ko pdhaa rhi thi Yash mujhe kuch btana tha Something..? Hyyy |
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| 16229. |
Traingle koi mujhe aache se samajha skta hai plz koi samja do yrr |
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Answer» Are bhai kuch v samajh me n aata hai proof karne wa la hi samajha do Ayush ????****??? Kya h Kya smjhna hai triangle chapter mein? |
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| 16230. |
1/sec theta -1 +1/sec theta +1=2 cot |
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| 16231. |
Find the square square root of 226 |
| Answer» 15.033296378372 | |
| 16232. |
Find K if kx (x-3)+8=0 has real and equal roots |
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Answer» Kx²-3x+8=0Given condition : real and equal rootsie.,D=0 or b²-4ac =0a=k, b=-3kc=8b²-4ac=0[(-3k)² - 4ac]=0[9k²-32k] =09k²-32k=09k²=32k9k*k=32kK=32k/9kk=32/9or k=3.555..... approximately: 3.6 X=12 therefore k=6 |
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| 16233. |
How many chapters come in exam |
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Answer» All Atleast mention which exam All chapters of mathematics will come |
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| 16234. |
Periodic test question paper |
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| 16235. |
In an A.P, the nth term is 1/m and the mth term is 1/n. find (mn)th term ,sum of first (mn) terms. |
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| 16236. |
Factories 4u² +8u |
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Answer» p(x) = 4u²+8u\xa04u ( u + 2 ) = 0[4u + 0 ] [ u + 2 ] ------> factorized formhence ,u = 0u = -2Here ,a = 4b = 8c = 0Relation between zeros and its coefficients :-Sum of zeroes = 0 + -2 = - 2\xa0= -b/a = -8/4 = -2Product of zeroes = 0 x -2 = 0\xa0= c/a = 0/4 = 0\xa0 4u{u+2} 2u[2u+4] 4u (u+2) |
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| 16237. |
What is. Quadratic formula |
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Answer» -b+-root bsquar -4ac/2a Formula is -b +- root b square -4ac/2a Yfiiohubj -b +- root bsquare + 4ac-------------------------------------- 2a |
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| 16238. |
Which of you are taking standard maths and which one basic maths |
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Answer» Basic Standard.... |
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| 16239. |
So that sequence diffine by an=3n²-5 is not ap |
| Answer» I don\'t know | |
| 16240. |
Any change in cbse math syllabus |
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Answer» There is no change in Maths syllabus Dear Vikram@diti? No |
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| 16241. |
Write the value of 3 cot^2 theta - 3 cosec^2 theta |
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Answer» Take 3 common Then cot^ 2thetha -cosec^2thetha =-1Answer is -3 The value of this equation is 3 |
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| 16242. |
Solve for x and y x+y=5 and 2x-3y=4 |
| Answer» X+y=5. 2x-3y=4. 3(x+y=5) = 3x+3y=15. 3x+3y=152x-3y=4. X=19/5 . y=6/5 | |
| 16243. |
1.8*7.2\\5.4 |
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Answer» 2.4 2.4 2.4 |
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| 16244. |
Solve quadratic equation by factorization method one by x minus one by x minus two |
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| 16245. |
(x+1)/(x-1)+(x-2)/(x+2)=4-(2x+3)/(x-2) find values of x |
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Answer» Question:\xa0X+1/x-2 +x-2/x+2=4-2x+3/x-2Answer:x = Step-by-step explanation:Compare above equation withax²+bx+c=0,we geta=5,b=19,c=-30Discreminant (D)=b²-4ac=(19)²-4×19×(-30)=361+600=961ByQuadratic Formula:x=[-b±√D]/(2a)= =Now,x= Or x =x = x = X=2/5 , x=-1 |
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| 16246. |
Find the value of(3tan 41/cot 49)241 |
| Answer» Full question:\xa0Find the value of: (3 tan 41/cot 49)^2 - (sin35sec55/tan10tan20tan60tan70tan80)^2Answer:26/3Step-by-step explanation:Given find the value of:(3 tan 41/cot 49)^2 - (sin35 sec55 / tan10 tan20 tan60 tan70 tan80)^2(3 tan 41 / cot (90 - 41) - sin 35 sec(90 - 35) / tan 10. tan 20. tan 60 . tan(90 - 20) tan(90 - 10)(3 tan 41 / tan 41)^2 - (sin 35 sec(90 - 35) / tan 10 tan 20 tan 60 tan(90 - 20)tan(90 - 10))^2(3 x 1)^2 - (sin 35 cosec 35 / tan 10. cot 10. tan 20. cot 20. tan 60)^29 - (1 / 1 x 1 x √3)^2 (because cosec 35 = 1/sin35)9 - 1 / 327 - 1 / 3\xa026 / 3 | |
| 16247. |
How to factorize the big number like x**2+5x+1800 in easiest way |
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Answer» Method It can done by hit and trial nethod |
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| 16248. |
What is the formula of a+b whole square? |
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Answer» (A+B)^2=a^2+b^2+2ab@diti? A^2+B^2+2AB a²+2ab+b² |
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| 16249. |
(3+1)(3power+1)(3power4+1)(3power8+1)(3power16+1)= |
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Answer» What do you mean by 3power+1...? Something is wrong in this question ? Hlwww |
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| 16250. |
√1-sin A /1+sin A= cosA/1+sin A |
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Answer» Mayank g thank u Is it √1-sinA or 1-sinA Try to find it on google Solve LHS and RHS separately so you will get (cosecA-1) in both sides. Someone tell Pls it\'s urgent |
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